CODEWITHSUNDEEP
rstringr
Szymon Fraś
Szymon Fraś

Reputation: 106

extract string before/after character without this character in r

I have a strings that look like that:

strings <- c("100% 16 Feedback Kakkanad, Ernakulam","98% 159 Feedback Greater Kailash Part 1, Delhi","Bannerghatta Road, Bangalore ₹250 Available on Su")

I want to extract number of feedbacks using stringr. After some tries I almost got it.

stringr
str_extract_all(strings,"\\%.*[0-9]\\s", simplify = T)

Function above returns vector:

"% 16 ", "% 159 ",""

I would like to extract only this number, so remove % and white spaces. I tried to search, but I didn't know how to define my problem in fewer words :(

Upvotes: 1

Views: 365

Answers (1)

akrun
akrun

Reputation: 887118

We can use str_extract with a regex lookaround to extract the digits that follow any zero or more spaces (\\s*) followed by the string 'Feedback'

library(stringr)
as.numeric(c(str_extract_all(strings, "\\d+(?=\\s*Feedback)", simplify = TRUE)))
#[1]  16 159

The issue with matching the % is that it also become part of the output. Instead, it can be a regex lookaround

as.numeric(str_extract_all(strings,"(?<=\\%\\s)[0-9]+", simplify = TRUE))
#[1]  16 159

Upvotes: 1

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